Find All Complex Solutions 8cos(theta)^2=2

$8 \cos^{2} (θ) = 2$

Step 1

Subtract $2$ from both sides of the equation.

$8 \cos^{2} (θ) - 2 = 0$

Step 2

Factor the left side of the equation.

Factor $2$ out of $8 \cos^{2} (θ) - 2$ .

Factor $2$ out of $8 \cos^{2} (θ)$ .

$2 (4 \cos^{2} (θ)) - 2 = 0$

Factor $2$ out of $- 2$ .

$2 (4 \cos^{2} (θ)) + 2 (- 1) = 0$

Factor $2$ out of $2 (4 \cos^{2} (θ)) + 2 (- 1)$ .

$2 (4 \cos^{2} (θ) - 1) = 0$

Rewrite $4 \cos^{2} (θ)$ as ${(2 \cos (θ))}^{2}$ .

$2 ({(2 \cos (θ))}^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$2 ({(2 \cos (θ))}^{2} - 1^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 \cos (θ)$ and $b = 1$ .

$2 ((2 \cos (θ) + 1) (2 \cos (θ) - 1)) = 0$

Remove unnecessary parentheses.

$2 (2 \cos (θ) + 1) (2 \cos (θ) - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 \cos (θ) + 1 = 0$

$2 \cos (θ) - 1 = 0$

Step 4

Set $2 \cos (θ) + 1$ equal to $0$ and solve for $θ$ .

Set $2 \cos (θ) + 1$ equal to $0$ .

$2 \cos (θ) + 1 = 0$

Solve $2 \cos (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$2 \cos (θ) = - 1$

Divide each term in $2 \cos (θ) = - 1$ by $2$ and simplify.

Divide each term in $2 \cos (θ) = - 1$ by $2$ .

$\frac{2 \cos (θ)}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cos (θ)}{2} = \frac{- 1}{2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$\cos (θ) = - \frac{1}{2}$

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (- \frac{1}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$θ = \frac{2 π}{3}$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 5

Set $2 \cos (θ) - 1$ equal to $0$ and solve for $θ$ .

Set $2 \cos (θ) - 1$ equal to $0$ .

$2 \cos (θ) - 1 = 0$

Solve $2 \cos (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$2 \cos (θ) = 1$

Divide each term in $2 \cos (θ) = 1$ by $2$ and simplify.

Divide each term in $2 \cos (θ) = 1$ by $2$ .

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{2}$

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (\frac{1}{2})$

Simplify the right side.

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{3}$

Simplify $2 π - \frac{π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 6

The final solution is all the values that make $2 (2 \cos (θ) + 1) (2 \cos (θ) - 1) = 0$ true.

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Find All Complex Solutions 8cos(theta)^2=2

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